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Question

$$ABCD$$ is a rectangle in which $$AB=2BC$$. If $$M$$ is the mid-point of $$BC$$ and the  diagonal $$BD$$ meets $$AM$$ at $$O,$$ prove that
$$\triangle BOM \sim \triangle AOD$$
1297573_5724138bce20400e9714dda7847c6660.png


Solution

Ref.Image.
AB = 2 BC ... (given) 
In $$ \Delta $$ BOM and $$\Delta $$ AOD
$$\angle $$ BOM = $$\angle $$AOD  ... (vertically  opposite angles)
$$\angle $$ OBM = $$\angle $$ ODA .... (ultemate  angles)
$$ \Delta $$ BOM $$\sim $$ $$\Delta $$ AOD   ... [AA test of similarity ] 

1233830_1297573_ans_92a14f85fc334db7a73faf20447ef8ee.jpg

Mathematics

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